Differentiate $\sin ^2 x$ with respect to $\cos ^2 x$.

If $x = \frac{t^2}{1+t^5}$ and $y = \frac{2t^3}{1+t^5}$ where $t \neq -1$ is a parameter, then find $\frac{dy}{dx}$.

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then:

If $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$,then at $t = \frac{\pi}{4}$,find $\frac{dy}{dx}$.

If the parametric equations of a curve are given by $x = \cos \theta + \log \tan \frac{\theta}{2}$ and $y = \sin \theta$,then the points for which $\frac{dy}{dx} = 0$ are given by

The position of a point at time $t$ is given by $x = a + bt - ct^2$ and $y = at + bt^2$. Its acceleration at time $t$ is: